drawing graph: -x^4 / 6 + x^3 - x^2 / 2 - 13x / 3 + 4

[Nemo]

Hi,

I have a similar problem similar to this one and my questions are:

As per Subhotosh Khan
"f(x) = A(x+2)(x-1)(x-3)(x-4) → f(0) = A * 2 * (-1) * (-3) *(-4) → 4 = A * (-24) → A = ?" Does A equal -1/6?


​Therefore my equation to graph would be -1/6(x+2)(x-1)(x-3)(x-4) expanded to \(\displaystyle \, -\dfrac{x^4}{6}\, +\, x^3\, -\, \dfrac{x^2}{2}\, -\, \dfrac{13x}{3}\, +\, 4\)

Or am I missing something?

 

[stapel]

I have a similar problem similar to this one...

In that thread, the relevant portion of the original question was:

Let f be a quartic polynomial (ie. a polynomial of degree 4). Suppose that f has zeros at −2, 1, 3, 4 and that f(0) = 4

If f(x) is written in the form f(x) = A(x− a)(x− b)(x− c)(x− d) ,then find the values of A, a, b, c, d.

...and my questions are:

As per Subhotosh Khan
"f(x) = A(x+2)(x-1)(x-3)(x-4) → f(0) = A * 2 * (-1) * (-3) *(-4) → 4 = A * (-24) → A = ?" Does A equal -1/6?

Yes.

​Therefore my equation to graph would be -1/6(x+2)(x-1)(x-3)(x-4) expanded to \(\displaystyle \, -\dfrac{x^4}{6}\, +\, x^3\, -\, \dfrac{x^2}{2}\, -\, \dfrac{13x}{3}\, +\, 4\)

Assuming the expansion was done correctly, yes, this is what you'll be graphing.

Note: The factored form may be more helpful, since the zeroes are very clearly indicated. You'd start your graph by plotting the intercepts at x = -2, 1, 3, and 4.

 

[Nemo]

Thanks

Thanks it has help a lot.